The ionization constant for water\(({K_w})\) is\(2.9 \times 1{0^{ - 14}}\;at\;4{0^o}C\). Calculate\(\left( {{H_3}{O^ + }} \right),\left( {O{H^ - }} \right),pH,\;and\;pOH\) for pure water at \(4{0^o}C\).

The formula for\(pH\)is\(pH = - \log \frac{{c\left( {{H_3}{O^ + }} \right)}}{{mol{L^{ - 1}}}}\).

The formula for\(pOH\)is\(pOH = - \log \frac{{c\left( {O{H^ - }} \right)}}{{mol{L^{ - 1}}}}\).

Let, the ionization constant for water\(({K_w})\) is\(2.9 \times {10^{ - 14}}{\rm{\;at\;}}{40^ \circ }{\rm{C}}\).

Hence, the equation of water ionization is,

\({{\rm{H}}_2}{\rm{O}}(l) + {{\rm{H}}_2}{\rm{O}}(l) \to {{\rm{H}}_3}{{\rm{O}}^ + }(aq) + {\rm{O}}{{\rm{H}}^ - }(aq)\)

Thus, the ionization constant is,

\({K_w} = c\left( {{H_3}{O^ + }} \right) \cdot c\left( {O{H^ - }} \right)\)

Since, the concentration of\({{\rm{H}}_3}{{\rm{O}}^ + }{\rm{and\;O}}{{\rm{H}}^ - }\)are equal, it can be written as,

\(c\left( {{H_3}{O^ + }} \right) = c\left( {O{H^ - }} \right) = c\)

Calculate the concentration,

\(\begin{aligned}{2.9 \cdot {{10}^{ - 14}} = {c^2}}\\{c = \sqrt {2.9 \cdot {{10}^{ - 14}}} }\\{c = 1.7 \cdot {{10}^{ - 7}}\frac{{{\rm{mol}}}}{{\rm{L}}}}\end{aligned}\)

Therefore,

\(\begin{aligned}{c\left( {{H_3}{O^ + }} \right) = 1.7 \cdot {{10}^{ - 7}}\frac{{{\rm{mol}}}}{L}}\\{c\left( {O{H^ - }} \right) = 1.7 \cdot {{10}^{ - 7}}\frac{{{\rm{mol}}}}{L}}\end{aligned}\)

Find the \(pH\)and \(pOH\)of pure water,

Use the formula,

\(pH = - \log \frac{{c\left( {{H_3}{O^ + }} \right)}}{{mol{L^{ - 1}}}}\)

Substitute \(c\left( {{H_3}{O^ + }} \right) = 1.7 \cdot {10^{ - 7}}\frac{{{\rm{mol}}}}{L}\)

\(\begin{aligned}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,pH = - \log \frac{{1.7 \cdot {{10}^{ - 7}}mol{L^{ - 1}}}}{{mol{L^{ - 1}}}}}\\{pH = 6.8}\end{aligned}\)

Solve for\(pOH\),

\(pOH = - \log \frac{{c\left( {O{H^ - }} \right)}}{{mol{L^{ - 1}}}}\)

Substitute, \(c\left( {O{H^ - }} \right) = 1.7 \cdot {10^{ - 7}}\frac{{{\rm{mol}}}}{L}\)

\(pOH = - \log \frac{{1.7 \cdot {{10}^{ - 7}}mo{l^{ - 1}}}}{{mol{L^{ - 1}}}}\)

\(pOH = 6.8\)

Thus, the value of\(\left( {{{\rm{H}}_3}{{\rm{O}}^ + }} \right),\left( {{\rm{O}}{{\rm{H}}^ - }} \right),{\rm{pH}},{\rm{\;and\;pOH}}\) are,

\(c\left( {{H_3}{O^ + }} \right) = 1.7 \cdot {10^{ - 7}}\frac{{{\rm{mol}}}}{L}\)

\(c\left( {O{H^ - }} \right) = 1.7 \cdot {10^{ - 7}}\frac{{{\rm{mol}}}}{L}\)

\(\begin{aligned}pH = 6.8,\\pOH = 6.8\end{aligned}\)